Civil Engineering - Waste Water Engineering - Discussion

Discussion Forum : Waste Water Engineering - Section 3 (Q.No. 36)
36.
For the open drain (N = 0.025) shown in the below figure, the discharge is
26.88 cumecs
27.88 cumecs
28.88 cumecs
29.88 cumecs.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
17 comments Page 2 of 2.

Kruthi said:   6 years ago
How to calculate the area? Please tell me.

Haris khan said:   6 years ago
A= D (B+(B+2nD)/2).

Here D = 2
B = 1.6
n = 1( which is given as slope 1:1 we will take the base i.e horizontal)
So, Area = 7.2.

Dheeraj said:   6 years ago
It is very simple guys:

The formula for Area of trapezoidal tank =Bd +sd^2.
B - width.
d - depth.
S - slope(1vertical is to 1 horizontal i.e, =1).

So, Area is (1.6 x 2)+(2^2)=7.2.
So 7.2/.025 = 288.
(1)

ILA SAQAW said:   3 years ago
Area = 7.2 sqr-m.

Perimeter= 7.252 m.
Hydr. R = 0.99 say 1.

Assume the slope of 1 in 100.

V = 1/0.025 x (1) ^ 2/3 x (1/100)^ 1/2.
V = 4 m/sec.
Q = A x V.
Q = 7.2 Sqr-m * 4 m/sec,
Q= 28.8 cumec.
(2)

Gokul said:   3 years ago
2a+4b is equal to 7.2.
where a = 0.4.
b = 1.6.

Noorzaman said:   3 years ago
Anyone, Please explain me the velocity formula.

Pronay said:   3 years ago
@ILA SAQAW.

How perimeter is 7.2? Please explain.


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